Problem: $\overline{AC}$ is $10$ units long $\overline{BC}$ is $10$ units long $\overline{AB}$ is $10\sqrt{2}$ units long What is $\sin(\angle ABC)$ ? $A$ $C$ $B$ $10$ $10$ $10\sqrt{2}$
Solution: SOH CAH TOA in = pposite over ypotenuse opposite $= \overline{AC} = 10$ hypotenuse $= \overline{AB} = 10\sqrt{2}$ $\sin(\angle ABC)=\frac{10}{10\sqrt{2}}$ $=\dfrac{ \sqrt{2}}{2}$